I have managed to get my first python script to work which downloads a list of .ZIP files from a URL and then proceeds to extract the ZIP files and writes them to disk.
I am now at a loss to achieve the next step.
My primary goal is to download and extract the zip file and pass the contents (CSV data) via a TCP stream. I would prefer not to actually write any of the zip or extracted files to disk if I could get away with it.
Here is my current script which works but unfortunately has to write the files to disk.
import urllib, urllister
import zipfile
import urllib2
import os
import time
import pickle
# check for extraction directories existence
if not os.path.isdir('downloaded'):
os.makedirs('downloaded')
if not os.path.isdir('extracted'):
os.makedirs('extracted')
# open logfile for downloaded data and save to local variable
if os.path.isfile('downloaded.pickle'):
downloadedLog = pickle.load(open('downloaded.pickle'))
else:
downloadedLog = {'key':'value'}
# remove entries older than 5 days (to maintain speed)
# path of zip files
zipFileURL = "http://www.thewebserver.com/that/contains/a/directory/of/zip/files"
# retrieve list of URLs from the webservers
usock = urllib.urlopen(zipFileURL)
parser = urllister.URLLister()
parser.feed(usock.read())
usock.close()
parser.close()
# only parse urls
for url in parser.urls:
if "PUBLIC_P5MIN" in url:
# download the file
downloadURL = zipFileURL + url
outputFilename = "downloaded/" + url
# check if file already exists on disk
if url in downloadedLog or os.path.isfile(outputFilename):
print "Skipping " + downloadURL
continue
print "Downloading ",downloadURL
response = urllib2.urlopen(downloadURL)
zippedData = response.read()
# save data to disk
print "Saving to ",outputFilename
output = open(outputFilename,'wb')
output.write(zippedData)
output.close()
# extract the data
zfobj = zipfile.ZipFile(outputFilename)
for name in zfobj.namelist():
uncompressed = zfobj.read(name)
# save uncompressed data to disk
outputFilename = "extracted/" + name
print "Saving extracted file to ",outputFilename
output = open(outputFilename,'wb')
output.write(uncompressed)
output.close()
# send data via tcp stream
# file successfully downloaded and extracted store into local log and filesystem log
downloadedLog[url] = time.time();
pickle.dump(downloadedLog, open('downloaded.pickle', "wb" ))
Below is a code snippet I used to fetch zipped csv file, please have a look:
Python 2:
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
resp = urlopen("http://www.test.com/file.zip")
myzip = ZipFile(StringIO(resp.read()))
for line in myzip.open(file).readlines():
print line
Python 3:
from io import BytesIO
from zipfile import ZipFile
from urllib.request import urlopen
# or: requests.get(url).content
resp = urlopen("http://www.test.com/file.zip")
myzip = ZipFile(BytesIO(resp.read()))
for line in myzip.open(file).readlines():
print(line.decode('utf-8'))
Here file is a string. To get the actual string that you want to pass, you can use zipfile.namelist(). For instance,
resp = urlopen('http://mlg.ucd.ie/files/datasets/bbc.zip')
myzip = ZipFile(BytesIO(resp.read()))
myzip.namelist()
# ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']
My suggestion would be to use a StringIO object. They emulate files, but reside in memory. So you could do something like this:
# get_zip_data() gets a zip archive containing 'foo.txt', reading 'hey, foo'
import zipfile
from StringIO import StringIO
zipdata = StringIO()
zipdata.write(get_zip_data())
myzipfile = zipfile.ZipFile(zipdata)
foofile = myzipfile.open('foo.txt')
print foofile.read()
# output: "hey, foo"
Or more simply (apologies to Vishal):
myzipfile = zipfile.ZipFile(StringIO(get_zip_data()))
for name in myzipfile.namelist():
[ ... ]
In Python 3 use BytesIO instead of StringIO:
import zipfile
from io import BytesIO
filebytes = BytesIO(get_zip_data())
myzipfile = zipfile.ZipFile(filebytes)
for name in myzipfile.namelist():
[ ... ]
I'd like to offer an updated Python 3 version of Vishal's excellent answer, which was using Python 2, along with some explanation of the adaptations / changes, which may have been already mentioned.
from io import BytesIO
from zipfile import ZipFile
import urllib.request
url = urllib.request.urlopen("http://www.unece.org/fileadmin/DAM/cefact/locode/loc162txt.zip")
with ZipFile(BytesIO(url.read())) as my_zip_file:
for contained_file in my_zip_file.namelist():
# with open(("unzipped_and_read_" + contained_file + ".file"), "wb") as output:
for line in my_zip_file.open(contained_file).readlines():
print(line)
# output.write(line)
Necessary changes:
There's no StringIO module in Python 3 (it's been moved to io.StringIO). Instead, I use io.BytesIO]2, because we will be handling a bytestream -- Docs, also this thread.
urlopen:
"The legacy urllib.urlopen function from Python 2.6 and earlier has been discontinued; urllib.request.urlopen() corresponds to the old urllib2.urlopen.", Docs and this thread.
Note:
In Python 3, the printed output lines will look like so: b'some text'. This is expected, as they aren't strings - remember, we're reading a bytestream. Have a look at Dan04's excellent answer.
A few minor changes I made:
I use with ... as instead of zipfile = ... according to the Docs.
The script now uses .namelist() to cycle through all the files in the zip and print their contents.
I moved the creation of the ZipFile object into the with statement, although I'm not sure if that's better.
I added (and commented out) an option to write the bytestream to file (per file in the zip), in response to NumenorForLife's comment; it adds "unzipped_and_read_" to the beginning of the filename and a ".file" extension (I prefer not to use ".txt" for files with bytestrings). The indenting of the code will, of course, need to be adjusted if you want to use it.
Need to be careful here -- because we have a byte string, we use binary mode, so "wb"; I have a feeling that writing binary opens a can of worms anyway...
I am using an example file, the UN/LOCODE text archive:
What I didn't do:
NumenorForLife asked about saving the zip to disk. I'm not sure what he meant by it -- downloading the zip file? That's a different task; see Oleh Prypin's excellent answer.
Here's a way:
import urllib.request
import shutil
with urllib.request.urlopen("http://www.unece.org/fileadmin/DAM/cefact/locode/2015-2_UNLOCODE_SecretariatNotes.pdf") as response, open("downloaded_file.pdf", 'w') as out_file:
shutil.copyfileobj(response, out_file)
I'd like to add my Python3 answer for completeness:
from io import BytesIO
from zipfile import ZipFile
import requests
def get_zip(file_url):
url = requests.get(file_url)
zipfile = ZipFile(BytesIO(url.content))
files = [zipfile.open(file_name) for file_name in zipfile.namelist()]
return files.pop() if len(files) == 1 else files
write to a temporary file which resides in RAM
it turns out the tempfile module ( http://docs.python.org/library/tempfile.html ) has just the thing:
tempfile.SpooledTemporaryFile([max_size=0[,
mode='w+b'[, bufsize=-1[, suffix=''[,
prefix='tmp'[, dir=None]]]]]])
This
function operates exactly as
TemporaryFile() does, except that data
is spooled in memory until the file
size exceeds max_size, or until the
file’s fileno() method is called, at
which point the contents are written
to disk and operation proceeds as with
TemporaryFile().
The resulting file has one additional
method, rollover(), which causes the
file to roll over to an on-disk file
regardless of its size.
The returned object is a file-like
object whose _file attribute is either
a StringIO object or a true file
object, depending on whether
rollover() has been called. This
file-like object can be used in a with
statement, just like a normal file.
New in version 2.6.
or if you're lazy and you have a tmpfs-mounted /tmp on Linux, you can just make a file there, but you have to delete it yourself and deal with naming
Adding on to the other answers using requests:
# download from web
import requests
url = 'http://mlg.ucd.ie/files/datasets/bbc.zip'
content = requests.get(url)
# unzip the content
from io import BytesIO
from zipfile import ZipFile
f = ZipFile(BytesIO(content.content))
print(f.namelist())
# outputs ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']
Use help(f) to get more functions details for e.g. extractall() which extracts the contents in zip file which later can be used with with open.
All of these answers appear too bulky and long. Use requests to shorten the code, e.g.:
import requests, zipfile, io
r = requests.get(zip_file_url)
z = zipfile.ZipFile(io.BytesIO(r.content))
z.extractall("/path/to/directory")
Vishal's example, however great, confuses when it comes to the file name, and I do not see the merit of redefing 'zipfile'.
Here is my example that downloads a zip that contains some files, one of which is a csv file that I subsequently read into a pandas DataFrame:
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
import pandas
url = urlopen("https://www.federalreserve.gov/apps/mdrm/pdf/MDRM.zip")
zf = ZipFile(StringIO(url.read()))
for item in zf.namelist():
print("File in zip: "+ item)
# find the first matching csv file in the zip:
match = [s for s in zf.namelist() if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv(zf.open(match), low_memory=False, skiprows=[0])
(Note, I use Python 2.7.13)
This is the exact solution that worked for me. I just tweaked it a little bit for Python 3 version by removing StringIO and adding IO library
Python 3 Version
from io import BytesIO
from zipfile import ZipFile
import pandas
import requests
url = "https://www.nseindia.com/content/indices/mcwb_jun19.zip"
content = requests.get(url)
zf = ZipFile(BytesIO(content.content))
for item in zf.namelist():
print("File in zip: "+ item)
# find the first matching csv file in the zip:
match = [s for s in zf.namelist() if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv(zf.open(match), low_memory=False, skiprows=[0])
It wasn't obvious in Vishal's answer what the file name was supposed to be in cases where there is no file on disk. I've modified his answer to work without modification for most needs.
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
def unzip_string(zipped_string):
unzipped_string = ''
zipfile = ZipFile(StringIO(zipped_string))
for name in zipfile.namelist():
unzipped_string += zipfile.open(name).read()
return unzipped_string
Use the zipfile module. To extract a file from a URL, you'll need to wrap the result of a urlopen call in a BytesIO object. This is because the result of a web request returned by urlopen doesn't support seeking:
from urllib.request import urlopen
from io import BytesIO
from zipfile import ZipFile
zip_url = 'http://example.com/my_file.zip'
with urlopen(zip_url) as f:
with BytesIO(f.read()) as b, ZipFile(b) as myzipfile:
foofile = myzipfile.open('foo.txt')
print(foofile.read())
If you already have the file downloaded locally, you don't need BytesIO, just open it in binary mode and pass to ZipFile directly:
from zipfile import ZipFile
zip_filename = 'my_file.zip'
with open(zip_filename, 'rb') as f:
with ZipFile(f) as myzipfile:
foofile = myzipfile.open('foo.txt')
print(foofile.read().decode('utf-8'))
Again, note that you have to open the file in binary ('rb') mode, not as text or you'll get a zipfile.BadZipFile: File is not a zip file error.
It's good practice to use all these things as context managers with the with statement, so that they'll be closed properly.
So an update, I found my compile issue was that I needed to change my notebook to a py file and choosing save as doesn't do that. So I had to run a different script turn my notebook to a py file. And part of my exe issue was I was using the fopen command that apparently isn't useable when compiled into a exe. So I redid the code to what is above. But now I get a write error when trying to run the script. I can not find anything on write functions with os is there somewhere else I should look?
Original code:
import requests
import json
import pandas as pd
import csv
from pathlib import Path
response = requests.get('url', headers={'CERT': 'cert'}, stream=True).json()
json2 = json.dumps(response)
f = open('data.json', 'r+')
f.write(json2)
f.close()
Path altered code:
import requests
import json
import pandas as pd
import csv
from pathlib import Path
response = requests.get('url', headers={'CERT': 'cert'}, stream=True).json()
json2 = json.dumps(response)
filename = 'data.json'
if '_MEIPASS2' in os.environ:
filename = os.path.join(os.environ['_MEIPASS2'], filename)
fd = open(filename, 'r+')
fd.write(json2)
fd.close()
The changes to the code allowed me to get past the fopen issue but created a write issue. Any ideas?
If you want to write to a file, you have to open it as writable.
fd = open(filename, 'wb')
Although I don't know why you're opening it in binary if you're writing text.
I would like to convert a large batch of MS Word files into the plain text format. I have no idea how to do it in Python. I found the following code online. My path is local and all file names are like cx-xxx (i.e. c1-000, c1-001, c2-000, c2-001 etc.):
from docx import [name of file]
import io
import shutil
import os
def convertDocxToText(path):
for d in os.listdir(path):
fileExtension=d.split(".")[-1]
if fileExtension =="docx":
docxFilename = path + d
print(docxFilename)
document = Document(docxFilename)
textFilename = path + d.split(".")[0] + ".txt"
with io.open(textFilename,"c", encoding="utf-8") as textFile:
for para in document.paragraphs:
textFile.write(unicode(para.text))
path= "/home/python/resumes/"
convertDocxToText(path)
Convert docx to txt with pypandoc:
import pypandoc
# Example file:
docxFilename = 'somefile.docx'
output = pypandoc.convert_file(docxFilename, 'plain', outputfile="somefile.txt")
assert output == ""
See the official documentation here:
https://pypi.org/project/pypandoc/
You can also use the library docx2txt in Python. Here's an example:
I use glob to iter over all DOCX files in the folder.
Note: I use a little list comprehension on the original name in order to re-use it in the TXT filename.
If there's anything I've forgotten to explain, tag me and I'll edit it in.
import docx2txt
import glob
directory = glob.glob('C:/folder_name/*.docx')
for file_name in directory:
with open(file_name, 'rb') as infile:
outfile = open(file_name[:-5]+'.txt', 'w', encoding='utf-8')
doc = docx2txt.process(infile)
outfile.write(doc)
outfile.close()
infile.close()
print("=========")
print("All done!")`
GroupDocs.Conversion Cloud SDK for Python supports 50+ file formats conversion. Its free plan provides 150 free API calls monthly.
# Import module
import groupdocs_conversion_cloud
from shutil import copyfile
# Get your client_id and client_key at https://dashboard.groupdocs.cloud (free registration is required).
client_id = "xxxxx-xxxx-xxxx-xxxx-xxxxxxxx"
client_key = "xxxxxxxxxxxxxxxxxxxxxxxxxxxx"
# Create instance of the API
convert_api = groupdocs_conversion_cloud.ConvertApi.from_keys(client_id, client_key)
try:
#Convert DOCX to txt
# Prepare request
request = groupdocs_conversion_cloud.ConvertDocumentDirectRequest("txt", "C:/Temp/sample.docx")
# Convert
result = convert_api.convert_document_direct(request)
copyfile(result, 'C:/Temp/sample.txt')
except groupdocs_conversion_cloud.ApiException as e:
print("Exception when calling get_supported_conversion_types: {0}".format(e.message))
I am trying to download and extract zip files using multiprocessing.Pool.But every time I execute the script only 3 zips will be downloaded and remaining files are not seen in the directory(CPU % is also touching 100%). Can someone help me how to solve this problem/suggest better approach and following the snippet that I have tried. I am completely new to multiprocessing. My goal is to download multiple files in parallel without reaching max CPU.
import StringIO
import os
import sys
import zipfile
from multiprocessing import Pool, cpu_count
import requests
filePath = os.path.dirname(os.path.abspath(__file__))
print("filePath is %s " % filePath)
sys.path.append(filePath)
url = ["http://mlg.ucd.ie/files/datasets/multiview_data_20130124.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/bbcsport.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/3sources.zip"]
def download_zips(url):
file_name = url.split("/")[-1]
response = requests.get(url)
sourceZip = zipfile.ZipFile(StringIO.StringIO(response.content))
print("\n Downloaded {} ".format(file_name))
sourceZip.extractall(filePath)
print("extracted {} \n".format(file_name))
sourceZip.close()
if __name__ == "__main__":
print("There are {} CPUs on this machine ".format(cpu_count()))
pool = Pool(cpu_count())
results = pool.map(download_zips, url)
pool.close()
pool.join()
output below
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
There are 4 CPUs on this machine
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
filePath is C:\Users\Documents\GitHub\Python-Examples-Internet\multi_processing
Downloaded bbcsport.zip
extracted bbcsport.zip
Downloaded 3sources.zip
extracted 3sources.zip
Downloaded multiview_data_20130124.zip
Downloaded movielists_20130821.zip
Downloaded movielists_20130821.zip
extracted multiview_data_20130124.zip
extracted movielists_20130821.zip
extracted movielists_20130821.zip
I've made a few minor tweeks in your function and it works fine. Please note that:
the file ".../movielists_20130821.zip" appears on your list twice, so you're donwloading the same thing twice (maybe a typo?)
The files ".../multiview_data_20130124.zip", ".../movielists_20130821.zip" and ".../3sources.zip", when extracted, yield a new directory. The file ".../bbcsport.zip", though, when extracted, places it's files in the root folder, your current working directory (see image below). Maybe you missed this check?
I added a try/except block in the donwload function. Why? Multiprocessing works by creating new (sub)processes to run stuff. If a subprocess throws an exception, the parent process does not catch it. So if any erros occour in this subprocess, it must be logged/handled there.
import sys, os
import zipfile
import requests
from multiprocessing import Pool, cpu_count
from functools import partial
from io import BytesIO
def download_zip(url, filePath):
try:
file_name = url.split("/")[-1]
response = requests.get(url)
sourceZip = zipfile.ZipFile(BytesIO(response.content))
print(" Downloaded {} ".format(file_name))
sourceZip.extractall(filePath)
print(" extracted {}".format(file_name))
sourceZip.close()
except Exception as e:
print(e)
if __name__ == "__main__":
filePath = os.path.dirname(os.path.abspath(__file__))
print("filePath is %s " % filePath)
# sys.path.append(filePath) # why do you need this?
urls = ["http://mlg.ucd.ie/files/datasets/multiview_data_20130124.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/bbcsport.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/3sources.zip"]
print("There are {} CPUs on this machine ".format(cpu_count()))
pool = Pool(cpu_count())
download_func = partial(download_zip, filePath = filePath)
results = pool.map(download_func, urls)
pool.close()
pool.join()
i suggest you do it using multithreading since it's an I/O bound like the following :
import requests, zipfile, io
import concurrent.futures
urls = ["http://mlg.ucd.ie/files/datasets/multiview_data_20130124.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/bbcsport.zip",
"http://mlg.ucd.ie/files/datasets/movielists_20130821.zip",
"http://mlg.ucd.ie/files/datasets/3sources.zip"]
def download_zips(url):
file_name = url.split("/")[-1]
response = requests.get(url)
sourceZip = zipfile.ZipFile(io.BytesIO(response.content))
print("\n Downloaded {} ".format(file_name))
sourceZip.extractall(filePath)
print("extracted {} \n".format(file_name))
sourceZip.close()
with concurrent.futures.ThreadPoolExecutor() as exector :
exector.map(download_zip, urls)
I am trying to download and unzip Kaggle dataset by python script(Python 3.5), but I get an error.
import io
from zipfile import ZipFile
import csv
import urllib.request
url = 'https://www.kaggle.com/c/quora-question-pairs/download/test.csv.zip'
response = urllib.request.urlopen(url)
c=ZipFile(io.BytesIO(response.read()))
After running this code, I get the following error.
BadZipFile: File is not a zip file
How can I get rid of this error? What's the cause?
Using requests module and some minor fix to http://ramhiser.com/2012/11/23/how-to-download-kaggle-data-with-python-and-requests-dot-py/ the solution is:
import io
from zipfile import ZipFile
import csv
import requests
# The direct link to the Kaggle data set
data_url = 'https://www.kaggle.com/c/quora-question-pairs/download/test.csv.zip'
# The local path where the data set is saved.
local_filename = "test.csv.zip"
# Kaggle Username and Password
kaggle_info = {'UserName': "my_username", 'Password': "my_password"}
# Attempts to download the CSV file. Gets rejected because we are not logged in.
r = requests.get(data_url)
# Login to Kaggle and retrieve the data.
r = requests.post(r.url, data = kaggle_info)
# Writes the data to a local file one chunk at a time.
f = open(local_filename, 'wb')
for chunk in r.iter_content(chunk_size = 512 * 1024): # Reads 512KB at a time into memory
if chunk: # filter out keep-alive new chunks
f.write(chunk)
f.close()
c = ZipFile(local_filename)